“And on the seventh day God finished His work which He had done, and He rested on the seventh day from all His work which He had done.”

(Genesis 2:2)

Chapter 1

INTRODUCTION

1.1 Historical perspective

1.2 The Rutherford scattering formula

1.3 The properties of the Rutherford differential cross-section

1.4 The experiments of Rutherford and his colleagues

1.5 Examination of the assumptions

1.6 The nuclear constituents

1.7 What is coming?

§ 1.1 Historical perspective

1895 The discovery of X-rays Röntgen1896 Discovery of radioactivity Becquerel

HeThU 42

23490

23892

1897 The discovery of electron J.J. Thomson1900 The discovery of the black body radiation formula Max Planck1905 The development of the theory of special relativity Albert Einstein1911 Rutherford’s atomic model Rutherford1912 Discovery of isotopes J.J. Thomson1913 Bohr’s theory of the hydrogen atom Niels Bohr1919 Induced nuclear transmutation J.J. Thomson1920 The radii of a few heavy nuclei Chadwick ~ 10-14 m = 10 F << 10-10 m

1928 Alpha decay Gamow, Gurney and Condon1932 Discovery of neutron Chadwick1932 n-p hypothesis Heisenberg1932 Discovery of positron Anderson1934 Beta decay Enrico Fermi1935 Roles of mesons in nuclear forces Yukawa1936 Discovery of μ meson Anderson and Neddermeyer1946 Discovery of π meson Powell1956 Non-conservation of parity in beta decay Lee, Yang and 吳健雄教授

Atomic Physics

The physics of the electronic, extra-nuclear structure of atoms

Nuclear Physics

The physics of the atomic nucleus, believed to be constituted of neutrons and protons

Elementary Particle Physics

The physics of quarks and gluons, believed to be the constituents of protons and neutrons, and of leptons and gauge bosons and…who knows what else!

Quarks, gluons, leptons, and gauge bosons are believed to have no substructure.

SIMPLE NUCLEI

Hydrogen

Deuterium

Helium

Most visible mass in the universe is in the form of atomic nuclei

About 75% protons

The rest mainly helium

For every proton there are about 108 photons

Universe is essentially space and nuclei

Dark Energy and Dark Matter!

Big Bang !!

The universe is expanding→ BIG BANG model

a quick history

(0) at the end of the firstnanosecond (10-9 s) matterconsisted of quarks, leptonsand the various bosons whichtransmit the forces betweenthem

The fundamental particles

The fundamental particles

History of Nuclei

(1) As the universe expanded from very high density, pressure and temperatureit cooled allowing the strong force between the quarks (anti-quarks) to bindthem together into particles called hadrons (baryons or mesons).

(2) With further expansion the hadrons separated from each other, decaysoccurred and only the most stable species survived. At the end of the firstmicrosecond the only hadrons remaining were protons and neutrons togetherwith electrons (positrons), their neutrinos (antineutrinos) and photons.

History of Nuclei

(3) Up to first second the universe continued to expand and cool - radioactive decay and its inverse ensuring equilibrium between the numbers of protons and neutrons.(4) Up to the first fifteen minutes the temperature cooled sufficiently that neutrons and protons could bind together to form deuterons (one proton and one neutron) and the more tightly bound alpha particle (two protons and two neutrons). This was the epoch of primordial nucleosynthesis(5) For the next million years or so the universe was an expanding and slowly cooling mixture of nuclei, electrons, neutrinos and photons.

History of Nuclei

(6) By the time the universe was a thousand million years old, atoms and molecules had clustered together in large clouds which were further squeezed into stars by mutual gravitational attraction.(7) The collapse is halted and the star is fuelled by the process of nuclear fusion which converts hydrogen into helium and then into heavier nuclei. Thus a second phase of the production of nuclei occurred - so called stellar nucleosynthesis.

There are three types of radiations:

1. α-rays: These were found to be positively charged particles with a ratio of charge to mass about one half that of a singly charged hydrogen atom. It became clear afterward that these rays were energetic nuclei of helium.

2. β-rays: These are negatively charged particles which were found to be identical to the electrons found by J. J. Thomson.

3. γ-rays: These are electrically neutral particles with properties which identified them as energetic photons.

Three types of radiation

Displacement Law (by Russel, by Soddy and Fajans)

1. The emission of an α-particle reduces the atomic mass by 4 and the atomic number by 2.

2. The emission of a β-particle increases the atomic number by 1 and leaves the mass number unchanged.

α-rays

An α particle, or a helium nucleus, is composed of two protons and two neutrons, totally four nucleons. It carries +2e charge and has strong capability of ionization.

When α particles passing through materials high density of ions are created in a cylindrical shape along the path. It is called the “cylindrical ionization”.

High energy α particles lose their energy fast and can only travel through a short distance.

Ex. Po21084 (polonium) emits 5.3 MeV α-partilces

@ Alpha particles with 5.3 MeV is able to penetrate through 3.8 cm thickness of air.

@ A piece of paper of common thickness is able to stop those alpha particles.

@ They can not penetrate through human skin.

β-rays

Beta rays (electrons) interact with atoms. They lose energy by exciting or ionizing atoms to higher energy states and free states while traveling through materials.

Ion density created by electrons are far less than those from alpha particles with the same energy.

Electrons can travel through a much longer distance than alpha particles.

@ Electrons with 5.3 MeV is able to penetrate through 20 m thickness of air. This is roughly 500 times of alpha particles.

@ Electrons with 5.3 MeV is able to penetrate through 10 mm in Al and 2 mm in Pb .

@ β-rays are much more dangerous than α-rays .

γ-rays

Gamma rays are energetic photons. They interact with materials through three different kinds of effect.

1. Photoelectric effect Eγ < 0.4 MeV

2. Compton effect 0.4 MeV < Eγ < 5 MeV

3. Pair production effect Eγ > 5 MeV

@ These effects have very small effective cross section (very small probability of occurrence) thus huge traveling distance.

@ γ-rays are able to penetrate through ~ cm in Pb .

@ Watch out for γ-rays .

“And they heard the sound of Jehovah God walking about

in the garden in the cool of the day, and the man and his wife

hid themselves from the presence of Jehovah God

among the trees of the garden. “

(Genesis 3:8)

1. In 1906 Rutherford observed that a beam of α-particles became spread slightly on traversing a thin layer of material.

A layer of material insufficient to stop α-particles (e.g. gold 4μm thick) would scatter α-particles an average of about 9 degrees.

2. In 1909 Rutherford’s colleagues, Geiger and Marsden, observed that one in a few thousand α-particles suffered a scattering of greater than 90º.

Rare, Large-angle Scatters!!

Massive cores were encountered.

§ 1.2 The Rutherford scattering formula

Rutherford Scattering

Alpha particles hit atoms with massive nucleus cores.

1. The averaged 9 degrees small angle scattering is the result of many very small angle deflections (multiple scattering).

Multiple scattering: The deflection of the path is the result of the sum of many very small deflections in many atoms, all

uncorrelated.

2. The rare large angle scattering is the result of a single encounter (single scattering) with an atom.

Single scattering: The deflection of the path of a particle crossing a layer of material is the result of a significant deflection in one, and only one, encounter with

an atom.

Size of Nuclei

Atomic radius of aluminum = 1.3 x 10-10 m

Nuclear radius aluminum = 3.6 x 10-15 m

Size Comparison

Ernest Rutherford

(1871- 1937)

The Rutherford scattering formula

TZzep 02 4/

Assumptions

1 The nuclear model2 Target nucleus fixed (no recoil)3 Point-like charges4 Coulomb force only5 Elastic scattering6 Classical mechanics

p distance of closest approach for b = 0

The notation for quantities used in deriving Rutherford’s formula for the differential scattering cross-section for the elastic scattering of one charged particle by a fixed charged target particle.

m massv velocityT kinetic energyze electric charge

Incidentparticle

Ze charge of target nucleus (at O)b impact parameterd distance of closest approach (at D)u velocity of incident particle at Dθ angle of scatterr, φ polar coordinates with respect to OD of point (X) on the trajectory of particle.

Quantities in the figure

1. The orbit is hyperbolic and at D the incident particle is at its distance of closest approach, d.

2. The orbit is clearly symmetric about the line OD.

3. If b was zero the incident particles would approach to a distance p. At this point the incident kinetic energy is transformed into mechanical potential energy in the Coulomb field, therefore:

022 4/

2

1 Zzepmv

Step 1 To find the connection between b and θ.

In this system the angular momentum about O is conserved.

dt

dmrmvb

2

hence

vb

d

r

dt

2

(1)

(2)

Consider the component of the linear momentum in the direction OD.This changes from –mvsin(θ/2) to mvsin(θ/2).

)2/sin(2)]2/sin([)2/sin( mvmvmv

The total momentum change along the OD direction is

(3)

At X the rate of change of this momentum is the component of the Coulomb repulsion in the direction OD.

dtrZzemv

cos)4/(2

sin2 20

2 (4)

From equation (2) vb

drdt

2 put this into equation (4)

then we have

2/)(2/)(

2/)(

2/)(0

2

sin2

cos42

sin2

b

mvpd

vb

Zzemv (5)

b

p

22tan

Finally (6)

b

p

22tan

This equation means that as b decreases θ increases.

Step 2 To derive a first cross-section.

To suffer an angle of scatter greater than

the impact parameter b must be less than

)2/cot()2/( p

To suffer an angle of scatter greater than

the impact parameter b must be less than )2/cot()2/( p

This means the incident particle must strike a disk of this radius centered at O and perpendicular to the velocity v.

The area, σ , presented by the nucleus for scattering through an angle greater than is the area of this disk.

That is 2

cot4

)( 22

p (7)

Equation (7) can also be written as

2cot

44)( 2

2

0

2

T

Zze

(8)

The area σis called a cross-sectioncross-section.

Step 3 To obtain the angular differential cross-section.

We need the cross-section per unit solid angle located at an angle θ.

The element of solid angle dΩ between θand θ+dθ is given by

d sin 2 d

therefore

d

d

sin2

1

d

d

The dσ/dθwe need is )(d

d

from equation (8).

Hence we obtain

2

θcosec

16πd

d 4

2

0

2

Tε

Zze(9)

This is the famous Rutherford formula for the differential cross-section in Coulomb scattering.

d

d

θ

2

θcosec

16πd

d 4

2

0

2

Tε

Zze (9)

§ 1.3 The properties of the Rutherford differential cross-section

2

θcosec

16πd

d 4

2

0

2

Tε

Zze

The cross-section(1) decreases rapidly with increasing angle,θ,(2) becomes infinite at θ= 0,(3) is inversely proportional to the square of the incident particles kinetic energy, T,(4) is proportional to the square of the charge of the incident particle and of the target nucleus.

(9)

In the figure we have the momentum transfer q.

)2/sin(2 Pq (10)

1. Greater value of q means larger electric force.Larger electric force (or field) means close collisions therefore less cross-section with larger scattering angle θ.

2. At a fixed angle the required momentum transfer will increase as does T. Thus the cross-section must decrease with increasing T.

2

θcosec

16πd

d 4

2

0

2

Tε

Zze(9)

§ 1.4 Examination of the assumptions

1. Neglect of nuclear recoil: this can be avoid by transforming to the center-of-mass of the

collision. The formula is the same but the effective T is now the total kinetic energy in that frame and the angle of scatter and differential cross-section apply in that frame.

2. The classical approach to the orbit: a simple quantum mechanical approach using the Born

approximation gives the same answer.

3. The point-like charges: data from heavy nuclei (Au, Ag, and Cu targets) are all consistent

with Rutherford formula. Using targets of several light nuclei (i.e. Al) deviations from the Rutherford formula were found.

The effect of nuclear interaction at short distances!!

§ 1.4 Examination of the assumptions (continued)

4. Absence of other forces: Short range nuclear forces and magnetic effect (because of spin) are pres

ent.

5. Elastic scattering: The α-particles used by Rutherford were not sufficiently energetic to caus

e a significant number of inelastic collisions. By inelastic collision it means that one or both of the particles involved in the collision event become excited or disintegrate. After Rutherford, artificial sources of more energetic α-particles became available and these certainly can cause inelastic collisions.

6. Relativistic effect: If the incoming particles are energetic enough the relativistic effect is not

to be overlooked.

Example:

The scattering data of α-particle with energy exceeding 25MeV (Tα> 25MeV) on 92U deviate from what is predicted by the Rutherford formula.

The closest distance D in the head-on collision with Tα = 25MeV is

D = 10.6 F

When an α-particle comes near the 92U nucleus to within 10.6 F it barely touch the surface of the nucleus.

Energetic electron scattering experiments are able to probe internal structures of all nuclei.

§ 1.5 The nuclear constituents

Notation used to represent a given nuclide

NAZ XZ: atomic number

A: atomic mass number

N: neutron number

X: chemical symbol

HMM integer

M: the mass of a specific atom

MH: the mass of a hydrogen atom

NZA

From the previous discussion it might be thought that a nucleus is comprised of A protons and N electrons. This argument may well explain the β-radioactivity.

This is wrong!!

If we consider the nucleus of the helium atom. In this model there will be 4 protons and 2 electrons occupy a volume of half linear dimension of about 2 fermis.

From the uncertainty principle ~xp

For an electron confined in a region of 2 fermis its momentum is roughly

MeV 49~F 4

FMeV/c 3.197~

xp

Reason 1

MeV/c 49~F 4

FMeV/c 3.197~

xp

Electrons with kinetic energies of the order of 49 MeV should remain bound only if there is a potential well at least that deep. Such a potential well would have effects on the optical spectroscopy of helium which do not exist.

MeV 4949511.0 22220 pmE

MeV 49MeV 0.511 - MeV 490 mETe

where Te is electron’s kinetic energy

There is no electron in the nucleus!!

Reason 2

Another reason that it is impossible for electrons to stay in a nucleus.

When (A-Z) turns out to be an odd integer the model prediction of nuclei’s total angular momentum

does not agree with observations.

Example Consider the deuteron

121H A = 2, and Z = 1

Model: There should be 2 protons and an electron.

two protons one electron

2

1

2

1 2

1

2

3or

2

1

2

1

2

1

2

1

Possible spins of deuteron

But actually the deuteron spin is measured 1 J

Nomenclature

Nuclide A specific nuclear species, with a given proton number Z and neutron number N

Isotopes Nuclides of same Z and different NIsotones Nuclides of same N and different ZIsobars Nuclides of same mass number A (A = Z + N)Isomer Nuclide in an excited state with a measurable half-lifeNucleon Neutron or protonMesons Particles of mass between the electron mass (m0) and the proton mass (MH). The best-known mesons are π

mesons (≈ 270 m0), which play an important role in nuclear forces, and μ mesons (207 m0) which are important in cosmic-ray phenomena.Positron Positively charged electron of mass m0

Photon Quantum of electromagnetic radiation, commonly apparent as light, x ray, or gamma ray